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olga_2 [115]
3 years ago
13

You save $500 in a retirement account at age 25. it increases an average of 10% per year until you are 65. which formula and res

ult indicates how much money you will have in this account at age 65?
a.500 x (1 + 0.10) ^ (65 - 25) = $22,629.63b.500 x (1 + (0.10 x (65 - 25)) = $2,500 submit
Computers and Technology
1 answer:
Anvisha [2.4K]3 years ago
6 0
Any amount deposited in a bank at today will follow the compound interest formula for a period of time to multiply the amount of money deposited over time. Here too we can use the same formula: i.e

Amount after certain time = Principle * (1 + rate of interest/ number of installments in a year)^( number of installments in a year * time of deposit)

So, the correct formula for the amount at the age of 65 will be:

A = 500 * (1+0.1*/1)^(1*(65-25)) = 500 * (1+0.1/1)^(40*1) = <span>$22,629.63 </span>is the correct answer.
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Create a macro named mReadInt that reads a 16- or 32-bit signed integer from standard input and returns the value in an argument
timofeeve [1]

Answer:

;Macro mReadInt definition, which take two parameters

;one is the variable to save the number and other is the length

;of the number to read (2 for 16 bit and 4 for 32 bit) .

%macro mReadInt 2

mov eax,%2

cmp eax, "4"

je read2

cmp eax, "2"

je read1

read1:

mReadInt16 %1

cmp eax, "2"

je exitm

read2:

mReadInt32 %1

exitm:

xor eax, eax

%endmacro

;macro to read the 16 bit number, parameter is number variable

%macro mReadInt16 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;macro to read the 32 bit number, parameter is number variable

%macro mReadInt32 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;program to test the macro.

;data section, defining the user messages and lenths

section .data

userMsg db 'Please enter the 32 bit number: '

lenUserMsg equ $-userMsg

userMsg1 db 'Please enter the 16 bit number: '

lenUserMsg1 equ $-userMsg1

dispMsg db 'You have entered: '

lenDispMsg equ $-dispMsg

;.bss section to declare variables

section .bss

;num to read 32 bit number and num1 to rad 16-bit number

num resb 5

num1 resb 3

;.text section

section .text

;program start instruction

global _start

_start:

;Displaying the message to enter 32bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg

mov edx, lenUserMsg

int 80h

;calling the micro to read the number

mReadInt num, 4

;Printing the display message

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Printing the 32-bit number

mov eax, 4

mov ebx, 1

mov ecx, num

mov edx, 4

int 80h

;displaying message to enter the 16 bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg1

mov edx, lenUserMsg1

int 80h

;macro call to read 16 bit number and to assign that number to num1

;mReadInt num1,2

;calling the display mesage function

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Displaying the 16-bit number

mov eax, 4

mov ebx, 1

mov ecx, num1

mov edx, 2

int 80h

;exit from the loop

mov eax, 1

mov ebx, 0

int 80h

Explanation:

For an assembly code/language that has the conditions given in the question, the program that tests the macro, passing it operands of various sizes is given below;

;Macro mReadInt definition, which take two parameters

;one is the variable to save the number and other is the length

;of the number to read (2 for 16 bit and 4 for 32 bit) .

%macro mReadInt 2

mov eax,%2

cmp eax, "4"

je read2

cmp eax, "2"

je read1

read1:

mReadInt16 %1

cmp eax, "2"

je exitm

read2:

mReadInt32 %1

exitm:

xor eax, eax

%endmacro

;macro to read the 16 bit number, parameter is number variable

%macro mReadInt16 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;macro to read the 32 bit number, parameter is number variable

%macro mReadInt32 1

mov eax, 3

mov ebx, 2

mov ecx, %1

mov edx, 5

int 80h

%endmacro

;program to test the macro.

;data section, defining the user messages and lenths

section .data

userMsg db 'Please enter the 32 bit number: '

lenUserMsg equ $-userMsg

userMsg1 db 'Please enter the 16 bit number: '

lenUserMsg1 equ $-userMsg1

dispMsg db 'You have entered: '

lenDispMsg equ $-dispMsg

;.bss section to declare variables

section .bss

;num to read 32 bit number and num1 to rad 16-bit number

num resb 5

num1 resb 3

;.text section

section .text

;program start instruction

global _start

_start:

;Displaying the message to enter 32bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg

mov edx, lenUserMsg

int 80h

;calling the micro to read the number

mReadInt num, 4

;Printing the display message

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Printing the 32-bit number

mov eax, 4

mov ebx, 1

mov ecx, num

mov edx, 4

int 80h

;displaying message to enter the 16 bit number

mov eax, 4

mov ebx, 1

mov ecx, userMsg1

mov edx, lenUserMsg1

int 80h

;macro call to read 16 bit number and to assign that number to num1

;mReadInt num1,2

;calling the display mesage function

mov eax, 4

mov ebx, 1

mov ecx, dispMsg

mov edx, lenDispMsg

int 80h

;Displaying the 16-bit number

mov eax, 4

mov ebx, 1

mov ecx, num1

mov edx, 2

int 80h

;exit from the loop

mov eax, 1

mov ebx, 0

int 80h

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You have spent $4,000 on liquor for your bar. Your bar sales have been $24,000. What is your cost of sales for liquor, expressed
amid [387]

Answer:

17%

Explanation:

Given parameters:

Cost price of liquor = $4000

Total sale = $24000

Unknown:

Percentage cost of sale for liquor = ?

Solution:

To find the percentage cost of sale;

       %Cost of sale of liquor = \frac{cost price of liquor}{Total cost of sale }   x 100

Input the variables;

       %Cost of sale of liquor = \frac{4000}{24000}  x 100 = 16.67% = 17%

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Answer:

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